Đáp án:

$\dfrac{\sqrt{x}-2}{\sqrt{x}+2}$

Giải thích các bước giải:

$\dfrac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}$    (ĐKXĐ: $x\ne1;\;x\ge0$)

$=\dfrac{3x-3\sqrt{x}-3}{x-\sqrt{x}+2\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}$

$=\dfrac{3x-3\sqrt{x}-3}{(\sqrt{x}-1)(\sqrt{x}+2)}-\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-1)(\sqrt{x}+2)}$

$=\dfrac{3x-3\sqrt{x}-3-x+1-x+4}{(\sqrt{x}-1)(\sqrt{x}+2)}$

$=\dfrac{x-3\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}$

$=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{(\sqrt{x}-1)(\sqrt{x}+2)}$

$=\dfrac{(\sqrt{x}-1)(\sqrt{x}-2)}{(\sqrt{x}-1)(\sqrt{x}+2)}$

$=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}$

Vậy biểu thức đã cho bằng $\dfrac{\sqrt{x}-2}{\sqrt{x}+2}$ với $x\ne1;\;x\ge0$.